Given:
128g sample of titanium
2808J of heat energy
specific heat of titanium is 0.523 J/ g °C.
Required:
Change in temperature
Solution:
This can be solved
through the equation H = mCpT
where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.
Plugging in the
values into the equation
H = mCpT
2808J = (128g) (0.523
J /g °C) T
T
= 41.9 °C
Answer:
I know that option F is correct
Explanation:
Answer:
the enantiomeric excess of the mixture is 40%
Explanation:
The computation of the enantiomeric excess of the mixture is shown below:
As we know that
Hence, the enantiomeric excess of the mixture is 40%
Applying the conservation of mass:
Mass during bombardment:
242 (curium atom) + 4 (alpha particle)
= 246
This must equal the mass after bombardment:
1 (neutron) + x (unknown product)
246 = 1 + x
x = 245
The product is Californium-245
